Given: Torque $\tau = 100\text{ N m}$ Moment of inertia $I = 300\text{ kg m}^2$ Initial angular velocity $\omega_0 = 0$ (since it starts from rest) Time $t = 3\text{ s}$

From Newton's second law for rotational motion, the angular acceleration $\alpha$ is given by: $\tau = I\alpha \implies \alpha = \frac{\tau}{I} = \frac{100}{300} = \frac{1}{3}\text{ rad/s}^2$

Using the first equation of rotational kinematics, the final angular velocity $\omega$ is: $\omega = \omega_0 + \alpha t$ $\omega = 0 + \left(\frac{1}{3}\right) \times 3 = 1\text{ rad/s}$.